Find the area inside the outer loop and the outside the inner loop of the limacon $r=6\cos(\theta)-3$

Question

I keep getting $$ \frac{3}{2}\left ( 4\pi -3\sqrt{3} \right ) $$

My teacher says this is wrong, and I don't know how to approach this.

Answer 1

Although I am not sure where you might have gone wrong, you might have selected the wrong bounds, or made an error in your integration, as your answer is reasonably close. However, let's start from the top.

An image of the limacon, which shouldn't be too hard to graph by hand, should help us visualize what is going on.

The area of a polar curve between angles $\alpha$ and $\beta$ is given by: $A = \frac{1}{2}\int_{\alpha}^{\beta}[f(\theta)]^2d\theta$

Now $r = 6cos(\theta)-3 = f(\theta)$.

Let's look at a few points to determine which direction we are traveling and what values of $\theta$ might be critical:

when $\theta = 0, f(\theta) = 3$, and when $\theta = \pi/3, f(\theta) = 0$

If we integrate f between $\alpha = 0$ and $\beta = \pi/3$, we are traversing half of the inner loop — if this isn't clear, try drawing lines from the origin to each value of f along this path — so the area of the inner loop is twice this integral (by symmetry):

Inner loop area = $2 \cdot \frac{1}{2}\int_{\alpha}^{\beta}f[\theta]^2d\theta = \int_{0}^{\pi/3}(6cos(\theta)-3)^2d\theta = \cdots = 9\pi +\frac{9\sqrt{3}}{2}-18\sqrt{3}$

Now when $\theta = \pi/2, f(\theta) = -3$ (putting us at the bottom half of the loop where the limacon crosses the negative y-axis). This is not quite where we want to be so we move on: when $\theta = \pi, f(\theta) = -9$, putting us at the part of the limacon with the largest radius, crossing the positive x-axis.

If we integrate f between $\alpha = $ and $\beta = \pi$, we are traversing half of the outer loop, so the the area of the outer loop is twice this integral (by symmetry):

Outer loop area = $2 \cdot \frac{1}{2}\int_{\alpha}^{\beta}f[\theta]^2d\theta = \int_{\pi/3}^{\pi}(6cos(\theta)-3)^2d\theta = \cdots = 18\pi -\frac{9\sqrt{3}}{2}+18\sqrt{3}$

The area between the loops is [Outer loop area] - [inner loop area] = $18\pi -\frac{9\sqrt{3}}{2}+18\sqrt{3} - (9\pi +\frac{9\sqrt{3}}{2}-18\sqrt{3}) = 9\pi - \frac{18\sqrt{3}}{2} + 36\sqrt{3} = 9\pi + 27\sqrt{3}$

Answer 2

Derek answered the question with a graph. Finding the start and the end of the loops makes the sketching of the graph and setting up the integrals easier. In fact we need only a thought graph. So, we solve $r=6\cos\theta-3=0$ which gives the angles $\pm\frac{\pi}{3}$ and $\pm\frac{5\pi}{3}.$ Then, we can imagine that the inner loop is parametrized by $-\frac{\pi}{3}\leq\theta\leq\frac{\pi}3$ and the outer loop is by $\frac{\pi}{3}\leq\theta\leq\frac{5\pi}{3}$. Hence, $$A_{inner}=\frac12\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(6-3\cos\theta)^2d\theta=\frac{18\pi-27\sqrt3}{2}$$ and $$A_{outer}=\frac12\int_{\frac{\pi}{3}}^{\frac{5\pi}{3}}(6-3\cos\theta)^2d\theta=\frac{36\pi+27\sqrt3}{2}$$ and $A=A_{outer}-A_{inner}=9\pi+27\sqrt3$.